At 27ºC a gas is compressed suddenly such that its pressure becomes (1/8) of original pressure. Final temperature will be (γ= 5/3)

450 K
300 K
–142ºC
327ºC

Solution

\(T_{1}^{\gamma }P_{1}^{1-\gamma }=T_{2}^{\gamma }P_{2}^{1-\gamma }\)

A gas has pressure P and volume V.It is now compressed adiabatically to 1/32 times the original volume. Given that(32)^1.4= 128, the final pressure is (γ= 1.4)

P/128
P/32
32 P
128 P

Solution

By what percentage should the pressure of a given mass of a gas be increased so as to decrease its volume by 10% at a constant temperature?

8.1 %
9.1 %
10.1 %
11.1 %

Solution

The efficiency of a Carnot engine operating with reservoir temperatures of 100ºC and –23ºC will be

\(\frac{100+23}{100}\)
\(\frac{100-23}{100}\)
\(\frac{100+23}{373}\)
\(\frac{100-23}{373}\)

Solution

A polyatomic gas (γ= 4/3) is compressed to 1/8th of its volume adiabatically. If its initial pressure is P₀, its new pressure will be

8P₀
16P₀
6P₀
2P₀

Solution

P₀V^4/ = P₁(^V⁄₈)^4/3 ⇒ P₁ =P₀ 8^4/3 = 16P₀

The temperature of5 moles of a gas which was held at constant volume was changed from 100º to 120ºC. The change in the internal energy of the gas was found to be 80joule,the total heat capacity of the gas at constant volume will be equal to

8 joule per K
0.8 joule per K
4.0 joule per K
0.4 joule per K

Solution

dU = nC_vdT or 80 = 5 × C_vv= 4.0 joule/K

A perfect gas goes from a state A to another state B by absorbing 8 × 10⁵ J of heat and doing 6.5 × 10⁵ J of external work. It is now transferred between the same two states in another process in which it absorbs 10⁵ J of heat. In the second process

work done by gas is 10⁵ J
work done on gas is 10⁵ J
work done by gas is 0.5 × 10⁵ J
work done on the gas is 0.5 × 10⁵ J

Solution

dU = dQ - dW = (8 × 10⁵ - 6.5 × 10⁵) = 1.5 × 10⁵ J

dW = dQ - dU == 10⁵ - 1.5 × 10⁵ = - 0.5 × 10⁵ J

– ve sign indicates that work done on the gas is 0.5 × 10⁵ J

A refrigerator works between 0ºC and 27ºC. Heat is to be removed from the refrigerated space at the rate of 50 kcal/minute, the power of the motor of the refrigerator is

0.346 kW
3.46 kW
34.6 kW
346 kW

Solution

A system changes from the state (P₁, V₁) to (P₂, V₂) as shown in the figure. What is the work done by the system?

7.5 × 10⁵ joule
7.5 × 10⁵ erg
12 × 10⁵ joule
6 × 10⁵ joule

Solution

A mass of ideal gas at pressure P is expanded isothermally to four times the original volume and then slowly compressed a diabatically to its original volume. Assuming γ to be 1.5, the new pressure of the gas is

Solution

Let P and V be the initial pressure and volume of ideal gas. After is othermal expansion, pressure is P/4. So volume is 4 V.

Let P₁ be the pressure after adiabatic compression.

Then

P₁ V^γ = (P / 4)(4 V)^γ P₁=(P / 4)(4)^3/2 = 2P

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